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Ta có : \(2n^3-6n^2-2n+n^2-3n-1-2n^3+1\)
=> \(-5n^2-5n=-5\left(n^2+n\right)\)Như vậy luôn chia hết cho 5 với mọi n
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Bạn xem lại đề. Thay $n=1$ thì biểu thức không chia hết cho 7 nhé.
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\(=3^3.3^n+3.3^n+2^3.2^n+2^2.2^n=\)
\(=3^n\left(3^3+3\right)+2^n\left(2^3+2^2\right)=30.3^n+12.2^n=\)
\(=6\left(5.3^n+2.2^n\right)⋮6\)
\(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)
\(=3^{n+1}\left(9+3\right)+2^{n+2}\left(8+4\right)\)
\(=12.3^{n+1}+12.2^{n+2}=12.\left(3^{n+1}+2^{n+2}\right)\)
mà 12⋮6
\(\Rightarrow12.\left(3^{n+1}+2^{n+2}\right)⋮6\Rightarrow dpcm\)
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ta có:10n-52n
=102.10n-2-52.52n-2
=100.10n-2-25.52n-2
=25.(4.10n-2-52n-2) chia hết cho 25(đpcm).
Vậy 10n-52nchia hết cho 25
10^n-5^2n
=(10^2*10^n-2)-(5^2*5^2n-2)
=(100*10^n-2)-(25*5^2n-2)
=25*{4*[10^n-2]-(5^2n-2)} chia het cho 25
=>dpcm
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3n+3-2.3n+2n+5-7.2n
=3n.33-2.3n+2n.32-7.2n
=3n.(33-2)+2n.(32-7)
=3n.25+2n.25=25.(3n+2n) luôn chia hết cho 25